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I'm working the proof of the Stone-Weierstrass Approximation theorem using probability theory from "A Second Course in Probability" by Ross and Pekoz. The statement of the theorem in the book omits UNIFORM from the limit, where as the statement seen in other books such as Rudin is UNIFOM. The proof below is slightly modified from the book. I would like to determine if the form of the proof admits uniform convergence. (The assignment of writing up a presentation of the proof is complete; this is for my own sake having noticed the difference in statements).

Theorem: Any continuous function $f$ defined on the interval $[0,1]$ can be expressed as a limit of polynomial functions.

Proof:

Let $X_k$, $k\in\mathbb{N}$ be a sequence of iid random variables such that $P(X_k=1)=t=1-P(X_k=0)$

Let $\bar{X_n}$ be the average of the first $n$ samples. By the strong law of large numbers $\bar{X_n}\xrightarrow{\textrm{a.s.}}t\implies \bar{X_n}\to_pt\implies \bar{X_n}\to_d t$.

Since $f$ is continuous on a closed interval it is bounded; thus since $X_n\to_d X\implies E[g(X_n)]\to E[g(X_n)]$

we have that $E[f(\bar{X_n})]\to f(t)$ but $n\bar{X_n}$ is a binomial $(n,t)$ random variable; thus

$E[f(\bar{X_n})]=\sum_{k=0}^nf\left(\frac{k}{n}\right)\binom{n}{k}t^k(1-t)^{n-k}$

where the right side is clearly a polynomial.

Question: Now, this proves convergence; but can we prove uniform convergence as in the typical analytic proofs. My impression is that Egorov's Theorem implies almost uniform convergence; though I'm not versed enough in measure theory to fully understand/appreciate the statement of the theorem. From there I might use continuity arguments. Otherwise I would have to show uniform convergence almost everywhere at which point I'm lost.

I have both Rudin's and both of Ross' Probability books to work from though I've only studied about Half of the first Rudin (up to Stone Weierstrass) and first semester probability. Any pointers & help would be appreciated.

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up vote 2 down vote accepted

For uniformity you need a bit more than the SLLN, namely that there is some kind of uniformity in $t$. Typically this is handled by computing also the variance, see this page.

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